A plane is capable of flying at 180 km/hr....(an advanced Calc problem!)?

Question

I *need* to figure this baby out, and I'm not having much luck...can you guys walk me through it in as much detail as you can muster? I can really, really use it! Here it is:

"A plane is capable of flying at a speed of 180 km/hr in still air. The pilot takes off from an airfield and heads due north according to the plane's compass. After 30 minutes of flight time, the pilot notices that, due to the wind, the plane has actually traveled 80km at an angle 5 degrees east of north.
a) What is the wind velocity?
b) In what direction should the pilot have headed to reach the intended destination?"

...is it just me or have you guys given me a few different answers? lol

I'm still a bit confused...im really appreciative of all u guys have said thus far, but can you guys perhaps be a bit clearer explaining every step?

Answer 1

This is trigonometry, not calculus.

Subtract the vector of still air flight from actual flight to get the wind vector. For ease of calculation the magnitude of the vectors should be for a one hour flight.

Wind = ActualPath - StillAir

We are talking about bearings so 0° is north and positive angles are in a clockwise direction from north.

N = 2*80cos(5°) - 180 = 160cos(5°) - 180 ≈ -20.608843831
E = 2*80sin(5°) - 0 = 160sin(5°) ≈ 13.94491884

Calculate the magnitude of the wind using the Pythagorean Theorem.

| w | = √(N² + E²) ≈ √[(-20.608843831)² + 13.94491884²]
| w | ≈ 24.883 km
__________

Use similar principles to work out Part b). The angle can be calculated with tanθ.

Answer 2

This question is all to do with vector addition.

Planes distancevector due north (length 90)
Resultant distance vector at 005degrees (length 80)
Wind distance vector added to planes vector.

Can be calculated either by scale drawing or cosine rule.

a)Let W = wind vector

Then

W^2=90^2+80^2-2(90x80)Cos5
W=12.44

This represents distace in 30mins so wind speed is 24.88km/h

b) What is his intended destination - is it 90km due north? If so then this must be the resultant vector to the addition of the wind + the planes vectors.

Answer 3

you have to represent this situation using vectors, its a triangle ABC.AB goes from south to north of length 90, which is the distance traveled in 30 min.AC is going in a direction of 5 degrees east of north,angle BAC is 5 degrees,this represents the actual distance traveled in 30 min.and its length is 80 then join BC.The vector BC represents the distance traveled by the wind in 30 min.call this side y The angle ABC is the direction of the wind.call this angle x.
now apply the two rules
1) sum of vectors in the east - direction = 0
80 sin 5 = y sin x
y = 80 sin5 / sin x
2) sum of vectors in the north direction = 0
90 - y cos x = 80 cos 5
now sub y from 1) in 2)
90 - (80 sin5 / sin x) cos x = 80 cos 5
90 - 80 cos 5 = (80 sin5 / sin x) cos x
10.3 = 80 sin 5 / tan x
tan x = 10.3 /6.97= 1,47
angle x = 55.91 degrees
sub in
y = 80 sin5 / sin x
y = 8.42 km
that means that the wind traveled 8.42 km in 30 min
speed of wind = 8.42 * 2 = 16.84 km / hr
in a direction of 55.91 degrees east of south
the pilot should have headed 5 degrees west of north

Answer 4

use cosine law

the speed of the plane +wind=80/0.5=160km/hr
a^2=b^2+c^2-2bccosa
a^2=160^2+180^2-2(160)(180)cos5
a=24.9km/hr
use sine law to find the angle
sin5/24.9=sinx/160
x=34.1 degree180-34.1=145.9degree east of north

b)siny/24.9=sin34.1/180
y=4.45 degree
the angle is 4.45degree west of north